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From Wikipedia:

... a particle's distribution function is a function of seven variables, $f ( x , y , z , t ; v_ x , v_y , v_ z )$, which gives the number of particles per unit volume in single-particle phase space. It is the number of particles per unit volume having approximately the velocity $( v_x , v_y , v_z)$ near the place $( x , y , z )$ and time $( t )$.

We have $\frac{dx}{dt}=v_x(t)$, $\frac{dy}{dt}=v_y(t)$ and $\frac{dz}{dt}=v_z(t)$. Therefore the explicit form of $f$ is actually $$ f ( x(t) , y(t) , z(t) , t ; v_ x(t) , v_y(t) , v_ z(t) ) $$ Is this a correct interpretation?

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    $\begingroup$ I would say no. The variables (x,y,z) are not time dependent and do not represent a path of some particle. The correct interpretation is as follows: consider a neighborhood of the point (x,y,z) at time t. Then the distribution of velocities of particles in that neighborhood at that time is given by f(x,y,z,t;vx,vy,vz) $\endgroup$
    – John Donne
    Commented Nov 22, 2017 at 13:22

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No this not correct. With the particle distribution function you are asking the question "What is the probability of finding a particle at this position, with this velocity at this time?"

One way to picture it is say you have a device that can detect the presence of particles and can in some way filter them so that only particles with a given velocity are detected. If you make a measurement at a particular place, time and velocity, the probability that you will find something is given by the particle distribution function. The parameters do not represent the position and velocity of the particle, which evolve in time, the represent where you make your measurement, which you are free to control independently by hand.

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